Math 2310 New

MULTIVARIABLE CALCULUS

Newcastle University School of Mathematical and Physical Sciences

MATH 2310 Calculus of Science and Engineering Multivariable Calculus Strand

Contents 1. 2. 3. 4. 5. 6. 7. 8.

Introduction Coordinate Systems Parametric Curves and Surfaces Double Integrals Triple Integrals Line and Surface Integrals Vector Fields Surface Integrals of Vector Fields

c 2003 Newcastle University, Australia. °

1 9 13 17 39 49 55 65

MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

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1. Introduction This strand provides an introduction to the methods and techniques of the differential and integral calculus for real (scalar) and vector valued functions of several variables. That is; functions, f : Rn −→ R and f : Rn −→ Rm , respectively, where typically, n and m equal 2, or 3. NOTE: In books, or typed notes, vectors and vector valued functions are often denoted by letters in bold face font. In handwritten work it is usual to indicate them by underlining the letter with a ‘squiggle’. e.g. f . ˜

In the first instance we will concentrate on scalar valued functions; for example, f : R2 −→ R, given by f (x, y) := x2 + y 2 . The work in MATH1110 (or MATH1210) and in MATH1120 (or MATH1220) on one variable calculus, vectors and vector geometry, and functions of two variables therefore provides an essential background for this strand. NOTE: We use := to mean ‘defined by’ , or ‘equals by definition’. 1.1. Functions of 2 Variables. The graph of a function f : R2 −→ R, © ª (x, y, z) ∈ R3 : z = f (x, y) is a surface in 3-dimensional Cartesian (or, x-y-z) space, whose height above the x-y plane at the point (x0 , y0 ) is f (x0 , y0 ). For example; when f (x, y) = x2 + y 2 the equation z = f (x, y) defines an ‘upward opening’ paraboloid with vertex at the origin (see figure 1). For any value of the constant c the relationship f (x, y) = c defines a level curve of the function f in the x-y plane; namely: © ª (x, y) ∈ R2 : f (x, y) = c It is the ‘vertical’ projection onto the x-y plane of the intersection of the surface z = f (x, y) and the ‘horizontal’ plane z = c. For f (x, y) := x2 + y 2 the level curves are circles centre the origin (see figure 2). NOTE: Observe that generically the level curves of the graph of a function will indeed be curves, but in degenerate cases, where the surface has horizontal flats, they may not be.

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z 50 40 30 20 10 -4 0 -2 4 2 x

0 -4

-2

0

2

y

4

Figure 1. The surface f (x, y) = x2 + y 2 The level curves corresponding to a sequence of equally spaced values of the constant c form a contour map for the surface z = f (x, y), and the level curves are referred to as contours or contour lines. When f (x, y) = x2 + y 2 a contour map of the surface z = f (x, y) consists of a family of equally spaced concentric circles centred about the origin. y -4

-2

-4

2

-2

4

x

2

4

Figure 2. A contour map of f (x, y) = x2 + y 2 with c = 2, 4, 6, .. Let i := (1, 0) and j := (0, 1) denote the unit vectors in the direction of the positive x - axis and y - axis respectively. We introduce the vector differential operator ∇ := i

∂ ∂ + j ∂x ∂y

∇ operating on a function f : R2 −→ R produces the vector valued gradient, or gradient vector of f at (x, y):

MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

∂f ∂f grad f := ∇f = i+ j= ∂x ∂y

µ

∂f ∂f , ∂x ∂y

3

¶ .

For example, when f (x, y) = x2 + y 2 we have, grad f = ∇f = 2xi + 2yj = (2x, 2y). NOTE: ∇ is often pronounced ‘del’. In writing grad f := ∇f = ∂f i + ∂f j we have ∂x ∂y ∂f reverted to the usual convention of putting scalar multipliers (in this case ∂x and ∂f ) ∂y in front of vectors (here i and j). It is difficult to adhere to this when, say, defining ∇. Recall that the directional derivative of f : R2 −→ R at the point (x0 , y0 ) in the direction of a unit vector u = (u, v) is, f (x0 + hu, y0 + hv) − f (x0 , y0 ) h→0 h d = f (x0 + hu, y0 + hv)(0) dh ∂f ∂f = (x0 , y0 )u + (x0 , y0 )v ∂x ∂y (by the chain rule)

Du f (x0 , y0 ) = lim

= ∇f (x0 , y0 ) · .u Thus, Du f = ∇f · u The number Du f (x0 , y0 ) represents the instantaneous rate of change of f as we move from (x0 , y0 ) in the direction u. Suppose that u is a unit vector in the direction of the tangent at p = (x0 , y0 ) to the level curve of the surface z = f (x, y) through p; that is, the curve f (x, y) = c := f (x0 , y0 ). Then, instantaneously as we move from p in the direction u we are following a curve along which f is constant and so its rate of change is 0; thus, Du f (x0 , y0 ) = ∇f (x0 , y0 ) · u = 0 So, ∇f (x0 , y0 ) is orthogonal to u; that is, it points in the direction of the normal to the level curve of z = f (x, y) through (x0 , y0 ). Further, the directional derivative of f at (x0 , y0 ) in the direction of ∇f (x0 , y0 ) is D

∇f (x0 ,y0 ) k∇f (x0 ,y0 )k

∇f (x0 , y0 ) k∇f (x0 , y0 )k = k∇f (x0 , y0 )k ≥ 0,

f (x0 , y0 ) = ∇f (x0 , y0 ) ·

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so ∇f points in the direction of increasing f (indeed, in the direction in which f is increasing most rapidly) √ NOTE: For a vector u = (u, v), kuk denotes its length (or norm); kuk := u2 + v 2 . Recall: kuk2 = u · u and u · v = kukkvk cos θ, where θ is the angle between the vectors u and v. If r = (x, y) is the position vector of a point on the tangent at p = (x0 , y0 ) to the level curve of z = f (x, y) through p, then (r − p) · ∇f (p) = 0, giving the cartesian equation for the tangent line to the level curve of z = f (x, y) through p. The normal line to the level curve f (x, y) = c at p is given parametrically by r = p + t∇f (p),

−∞ < t < ∞.

p Example 1.1. Let f (x, y) = 25 − x2 − y 2 . The domain of f is {(x, y) : x2 +y 2 ≤ 25}, the disk radius 5 centre the origin. The surface z = f (x, y) is an ‘upside down’ hemispherical bowl of radius 5 sitting over this disk (see figure 3).

z

5 4 3 2 1 0 4

-4 2

-2 0 x

0 2

-2 4

-4

Figure 3. The surface f (x, y) =

y

p 25 − x2 − y 2

The level curves√are concentric circles centre the origin. For example, the level curve through p = (−2, 5) is given by f (x, y) = 4; that is, the circle x2 + y 2 = 32 . ! Ã −y −x . ,p ∇f (x, y) = p 25 − x2 − y 2 25 − x2 − y 2

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√ √ Thus, ∇f (−2, 5) = (1/2, − 5/4). The tangent line to the level curve f (x, y) = 4 at p is therefore given by, ³ √ ´ ³ √ ´ (x, y) − (−2, 5) · 1/2, − 5/4 = 0, that is, the line

√ 1 5 9 x− y + = 0, 2 4 4 (see figure 4), while the normal line is given by √ x = −2 + t/2, y = 5(1 − t/4) . y 5 4 3 2 1 -5

-4

-3

-2

-1

0

1

2

3 x

-1 -2 -3

Figure 4. √ The level curve f (x, y) = 4 and its tangent line at the point p = (−2, 5)

1.2. Functions of 3 Variables. We now extend the previous ideas to functions of three variables; for example, f : R3 −→ R, given by f (x, y, z) = x2 + y 2 − z. Generally, for each value of the constant c the relation f (x, y, z) = c defines a surface in 3-dimensional Cartesian space, which by analogy with level curves, we refer to as a level surface of f . Example 1.2. For example, the level sufaces of f (x, y, z) = x2 + y 2 − z are a family (one for each value of c) of identically shaped upward opening paraboloids with vertices at (−c, 0, 0) (see figure 5).

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z 60 40 20 0 -20 -5 0

x

5

-2

-4

2

0

4

y

Figure 5. Some level surfaces of f (x, y, z) = x2 + y 2 − z For a function f : R3 −→ R of three variables, we define ∇ := i

∂ ∂ ∂ + j + k ∂x ∂y ∂z

and ∂f i+ ∂x µ ∂f = , ∂x

grad f := ∇f =

∂f ∂f j+ k ∂y ∂z ¶ ∂f ∂f , , ∂y ∂z

where now i := (1, 0, 0), j := (0, 1, 0) and k := (0, 0, 1) are unit vectors in the x, y and z directions respectively. The directional derivative of f in the direction u = (u, v, w) is Du f = ∇f · u =

∂f ∂f ∂f u+ v+ w. ∂x ∂y ∂z

For reasons similar to those for functions of two variables, ∇f (x0 , y0 , z0 ) is orthogonal to the tangent plane to the level surface f (x, y, z) = c := f (x0 , y0 , z0 ) at the point p = (x0 , y0 , z0 ) and points in the direction of increasing f . Thus, the equation for this tangent plane is (r − p) · ∇f (p) = 0, where now r = (x, y, z), while the normal line to the surface at p is given parametrically by r = p + t∇f (p), −∞ < t < ∞.

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Example 1.3. For f (x, y, z) = x2 + y 2 − z ∇f = 2xi + 2yj − k = (2x, 2y, −1). So, the tangent plane to the level surface x2 + y 2 − z = 1 at the point (1, 1, 1) is ((x, y, z) − (1, 1, 1)) · (2, 2, −1) = 0; that is, the plane 2x + 2y − z = 3 (see figure 6).

z

50 40 30 20 10 0

-10 -20 -5 0

x

5 -4

-2

0

2

4

y

Figure 6. The tangent plane to the level surface x2 + y 2 − z = 1 at the point (1, 1, 1)

Note: A surface of the form z = g(x, y) corresponds to the level surface f (x, y, z) = 0, where f (x, y, z) := g(x, y) − z. So tangent planes and normal lines to such surfaces can be determined using the above formulas. Indeed, the last example can be viewed as an example of this form, where g(x, y) = x2 + y 2 − 1.

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2. Coordinate Systems Curves and regions in R2 are often more conveniently expressed using a coordinate system other than a Cartesian (x-y ) one. 2.1. Polar Coordinates (r, θ). Polar coordinates are one such system. A point p is determined by specifying its directed distance r from an origin 0, also referred to as the pole, and the angle θ the line segment from 0 to p makes with a prescribed ray (half-line) eminating from 0 called the polar axis, usually draw horizontally, and taken to coincide with the positive x - axis of a Cartesian coordinate system. Angles measured anticlockwise from the polar axis are by convention positive. In order that the values of r and θ can vary continuously as p moves about the plane it is necessary to allow them to assume all possible real values. However, this means, that unlike Cartesian coordinates, each point is represented by many different pairs of polar coordinates. For example; the polar coordinates (1, 5π/4), (1, −3π/4), √ (1, 13π/4) √ and (−1, π/4) all represent the same point p with Cartesian coordinates (−/ 2, −1/ 2) (see figure 7).

5π 4

π 4

13π 4

O O

O

O −3π 4

(1, 13π 4 )

(1, −3π4 )

(1, 5π4 )

π

(−1, 4 )

Figure 7. Multiple representations of a point in polar coordinates. If we wish points to have unique polar coordinates, we must restrict them to take only principal values; for instance r > 0 and 0 ≤ θ < 2π The relationships: x = r cos θ,

y = r sin θ

and so,

x y , sin θ = r r allow the conversion from polar to Cartesian coordinates and vice versa. r 2 = x2 + y 2 ,

cos θ =

NOTE: tan θ = y/x is not sufficient to determine θ. Example 2.1. We see that the circle centre the origin and radius 5, with Cartesian equation x2 + y 2 = 25,

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has the much simpler polar equation r = 5. Similarly, r = 2 cos θ 2 2 represents the circle √ (x − 1)√ + y = 1. The region: x/ 3 ≤ y ≤ 3x and 4 ≤ x2 + y 2 ≤ 9 is conveniently specified using polar coordinates by, π π ≤θ≤ and 2 ≤ r ≤ 3. 6 3

We now introduce alternative coordinate systems similar to polars for R3 . 2.2. Cylindrical coordinates (r, θ, z). In the cylindrical coordinate system a point p in R3 is specified by the polar coordinates r and θ of its projection in the x-y -plane and its directed distance z from the x-y -plane. The Cartesian coordinates of p are therefore, x = r cos θ,

y = r sin θ,

z = z.

To convert from Cartesian to cylindrical coordinates we use r2 = x2 + y 2 , cos θ =

x y , sin θ = , z = z. r r

Cylindrical coordinates are useful in situations where there is symmetry about a line, taken to be the z - axis. For any positive constant c, the axis of symmetry for the cylinder with Cartesian equation x2 + y 2 = c2 is the z - axis. In cylindrical coordinates this cylinder has the simple equation r = c. Example 2.2. The equation, z 2 = k2 r2 ,

k a constant,

specifies the double cone obtained by rotating the line z = kx in the x-z plane about the z - axis. The ‘football’ with Cartesian equation (x2 + y 2 )/a2 + z 2 = 1,

(0 < a < 1),

is specified in cylindrical coordinates by 2

z =1−

³ r ´2 a

.

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z

p (ρ, θ, φ)

z ρ

φ

φ

O x

r

θ

y

y

x

Figure 8. Spherical Coordinates 2.3. Spherical coordinates (ρ, θ, φ). Spherical coordinates are particularly useful in situations where there is symmetry about a point, taken to be the origin. The position of a point p in R3 is specified by: ρ (≥ 0), its distance from the origin 0; θ, the same angle as in cylindrical coordinates (that is, the angle the projection of 0p onto the x-y plane makes with the x - axis) and φ (0 ≤ φ ≤ π), the angle 0p makes with the positive z axis. From the diagram (see figure 8) we see that, in terms of spherical coordinates, cylindriical coordinates of p are given by, r = ρ sin φ,

θ = θ,

z = ρ cos φ,

and so the Cartesian coordinates of p are x = ρ sin φ cos θ,

y = ρ sin φ sin θ,

z = ρ cos φ. To convert from Cartesian coordinates to spherical coordinates we can use: p ρ = x2 + y 2 + z 2 , then cos φ = z/ρ remembering that 0 ≤ φ ≤ π, and finally, y sin θ = ρ sin φ together with x cos θ = ρ sin φ to determine θ.

Ã

y

=p

!

x2 + y 2

à =p

x x2 + y 2

, !

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For admissible values of the constant c, in spherical coordinates the equation ρ=c determines a spherical surface of radius c and centre the origin. Similarly, θ=c describes a half plane eminating from the z - axis, while φ=c represents a half cone with vertex at the origin and axis the z - axis.

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3. Parametric Curves and Surfaces 3.1. Parametric Curves. A general curve in R2 , or R3 , can be respectively represented parametrically by r(t) = x(t)i + y(t)j = (x(t), y(t)), or r(t) = x(t)i + y(t)j + z(t)k = (x(t), y(t), z(t)), for some range of the parameter values t. For eample,

µ

¶ cos t sin t r(t) = √ , √ , t t for π/4 ≤ t ≤ 12π, describes the spiral pictured in figure 9.

³ Figure 9. r(t) =

cos √ t , sin √t t t

´ for π/4 ≤ t ≤ 12π

NOTE:We can eliminate the parameter t from x = x(t), y = y(t) and z = z(t) to produces two independent relations between x, y and z. These define two surfaces whose intersection is the curve specified parametrically by r(t). If r : R −→ R3 : t 7→ r(t) = (x(t), y(t), z(t)) is a vector valued function parametrically representing the curve C, then its derivative at the point p0 := r(t0 ) on the curve is given by, r(t0 + h) − r(t0 ) dr (t0 ) = r0 (t0 ) := lim h→0 dt h = (x0 (t0 ), y 0 (t0 ), z 0 (t0 )) provided the limit exists.

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z r(t + h) - r (t) h

r'(t) p

0

r (t) r (t + h) O y x Figure 10. The vector r0 (t0 ). From the diagram (see figure 10) we see that r0 (t0 ) is a vector parallel to the tangent to C at p0 pointing in the direction of increasing t. Thus, the tangent line to C at p0 has equation (x, y, z) = r(t0 ) + tr0 (t0 ) Example 3.1. For the elliptical hellix given by, r(t) = (sin t, 2 cos t, t) (See figure 11.) we have, r0 (t) = (cos t, −2 sin t, 1), and so the tangent line at the point r(π/2) = (1, 0, π/2) is given by x = 1,

y = −2t,

z = π/2 + t

3.2. Parametric Surfaces. A general surface in R3 may be described by a vector valued function of two parameters u and v, r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k = (x(u, v), y(u, v), z(u, v)) , where (u, v) range over some domain D in the u-v plane. For example r(u, v) = ui + 3 cos vj + 3 sin vk, where −∞ < u < ∞ and 0 ≤ v < 2π, represents an infinite cylindrical surface with axis the x - axis and radius 3. Indeed, we see that if (x, y, z) is on the surface, then x = u can take any real value, while y 2 + z 2 = 9 cos2 v + 9 sin2 v = 9.

MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

z

z

10

10

5

-4

0

-2

15

5

2

4

y -4

0

-2

-5

-5

-10

-10

x 2

4

Figure 11. Two views of the helix r(t) = (sin t, 2 cos t, t) together with its tangent line at (1, 0, π/2). As with curves the two partial derivatives, ru =

∂r ∂x ∂y ∂z = i+ j+ k ∂u µ ∂u ∂u ∂u ¶ ∂x ∂y ∂z = , , ∂u ∂u ∂u

and rv =

∂r ∂x ∂y ∂z = i+ j+ k ∂v µ ∂v ∂v ∂v ¶ ∂x ∂y ∂z = , , , ∂v ∂v ∂v

evaluated at (u0 , v0 ) are respectively parallel to tangents at p0 to the curves r(u, v0 ) and r(u0 , v), both of which lie on the surface. Thus, provided they are linearly independent, they span a plane parallel to the tangent plane to the surface at p0 . A vector equation for the tangent plane at p0 is therefore, (x, y, z) = p0 + tru (u0 , v0 ) + srv (u0 , v0 ), where t and s are parameters, which can take all real values. Evaluated at (u0 , v0 )

¯ ¯i ¯ ∂x n := ru × rv = ¯¯ ∂u ¯ ∂x ∂v

j ∂y ∂u ∂y ∂v

¯ k ¯¯ ∂z ¯ ∂u ¯ ∂z ¯ ∂v

is perpendicular to the tangent plane at p0 ; that is, parallel to the normal to the surface at p0 , so (x, y, z) = p0 + tn is an equation for the normal line to the surface at p0 .

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Note that the Cartesian equation for the tangent plane at p0 is given by ((x, y, z) − p0 ) · n = 0. Example 3.2. For the surface r(u, v) = (u2 , v 2 , u + v) we have

¯ ¯ ¯i j k¯¯ ¯ n = ¯¯2u 0 1 ¯¯ ¯ 0 2v 1 ¯ = (−2v, −2u, 4uv)

at p0 = r(1, 1) n = (−2, −2, 4) and so the normal line to the surface at the point p0 is given by p0 + tn = (1, 1, 2) + (−2, −2, 4)t = (1 − 2t, 1 − 2t, 2 + 4t) and the tangent plane is given by ((x, y, z) − p0 ) · n = 0 (x − 1, y − 1, z − 2) · (−2, −2, 4) = 0 x + y − 2z = −2 see figure 12.

z

x

y

Figure 12. The tangent plane to the surface r(u, v) = (u2 , v 2 , u + v) at the point p0 = (1, 1, 2)

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4. Double Integrals 4.1. Introduction. Recall, that when the limit exists, Z b m X f (x)dx := lim f (xi )∆x, m→∞

a

i=1

where for each m, the interval I = [a, b] is divided into m subintervals I1 , I2 , · · · , Im each of length ∆x = (b − a)/m, so Ii = [ai−1 , ai ] where ai := a + i(b − a)/m, and xi is the midpoint of Ii ; that is, xi = a + (i − 1/2)(b − a)/m (see figure 13).

y

∆x

{

} 0

a

x1

x2

xi _1

x3

f(xi)

xi

b

x

Figure 13. Approximating the area under a curve. This integral may be interpreted as the (signed) area of the region lying in the vertical strip a ≤ x ≤ b and between the curve y = f (x) and the x - axis. Areas above the x axis being counted as positive, and those below negative. Motivated by this, we seek to develop the double integral: ZZ f (x, y)dA, R

representing the (signed) volume of that part of the vertical column with horizontal cross-section R lying between the surface z = f (x, y) and the x-y plane (see figure 14). For simplicity we begin with the case when R is a rectangle. Suppose that f (x, y) is a scalar valued function of two variables whose domain contains the rectangular region R = [a, b] × [c, d] = {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d}. For any pair of natural numbers m, n, divide R into mn subrectangles by dividing [a, b] into m subintervals I1 , I2 , · · · , Im each of length ∆x = (b − a)/m and [c, d] into n subintervals J1 , J2 , · · · , Jn each of length ∆y = (d − c)/n. Thus, for i = 1, 2, · · · , m, Ii = [ai−1 , ai ],

where ai = a + i(b − a)/m

and similarly, for j = 1, 2, · · · , n, Jj = [cj−1 , cj ],

where cj = c + j(d − c)/n.

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z graph of f

O y R x

Figure 14. A vertical column with horizontal cross-section R lying between the surface z = f (x, y) and the x-y plane. NOTE: a0 = a and am = b (c0 = c and cn = d). Let xi and yj denote the midpoints of Ii and Jj respectively, so xi = a + (i − 1/2)(b − a)/m and yj = c + (j − 1/2)(d − c)/n. Then, each subrectangle, Rij = Ii × Jj , has area ∆A = ∆x∆y and centre at the point (xi , yj ). For each choice of i and j, f (xi , yj )∆A is the (signed) volume of the rectangular prism (‘box’) with base Rij and ‘height’f (xi , yj ) and (for ∆x and ∆y small enough; equivalently, m and n large enough) this is approximately the volume of the vertical rectangular column bounded by Rij in the x-y plane and the surface z = f (x, y) (see figure 15). NOTE: Prisms which sit above the x-y plane have positive volumes, while those which hang below it are assigned negative volumes. Adding up the volumes of all such prisms; X f (xi , yj )∆A, i = 1, 2, · · · , m j = 1, 2, · · · , n

therefore gives an approximation to the volume of that part of the vertical rectangular column with horizontal cross-section R lying between the surface z = f (x, y) and the x-y plane. When the limit exists, we define, ZZ f (x, y)dA R X := lim f (xi , yj )∆A m,n→∞

i = 1, 2, · · · , m j = 1, 2, · · · , n

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19

z

f(xi,yi)

O

y x

Rij

Figure 15. The rectangular prism with base Rij and ‘height’f (xi , yj ).

and interpret it as the (signed) volume of that part of the vertical rectangular column with horizontal cross-section R lying between the surface z = f (x, y) and the x-y plane, where the volume of those parts of the solid above the x-y plane contribute positively to the volume while those below contribute negatively. RR It can be shown that the double integral, R f (x, y)dA exists, whenever f (x, y) is continuous at all points of R, or only has discontinuities at a finite number of points or along a finite number of piece-wise smooth curves. It can also be shown that the existence of the limit and its value is independent of where the points (xi , yj ) are chosen in the subrectangles Rij . Thus, they need not be the centres, but could for example be the top right hand corners (ai , cj ). As with integrals of functions of a single variable, except in very simple cases, double integrals are difficult to calculate directly using the above definition. For functions of a single variable the Fundamental Theorem of Calculus provides a way around this, by converting the problem to that of finding antiderivatives. We seek similar methods for handling double integrals. 4.2. Iterated Integrals. In the approximation, X

f (xi , yj )∆A,

i = 1, 2, · · · , m j = 1, 2, · · · , n

for the volume lying between the rectangle R in the x-y plane and the surface z = f (x, y), the volumes of the individual prisms over each of the subrectangles can be added up in any order. One systematic way of doing this is to sum the volumes along each column

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of subrectangles, n X

f (xi , yj )∆A (for the i’th column)

j=1

and then total these column sums: Ã n m X X i=1

! f (xi , yj )∆A

j=1

This leads us to: ZZ f (x, y)dA R

à n m X X

= lim

m,n→∞

i=1

m,n→∞

= lim

m→∞

= lim

m→∞

i=1

m X i=1

! f (xi , yj )∆x∆y

j=1

à lim

n→∞

m µZ X

n X

f (xi , yj )∆y ∆x

j=1

¶

d

f (xi , y)dy ∆x ¶

d

=

!

c

i=1

Z b µZ

f (xi , yj )∆A

j=1

à n m X X

= lim

!

f (x, y)dy dx, a

c

where for the purpose of evaluating the inner integral x is regarded as a constant. This last expression is called an iterated integral, which we usually write with the brackets omitted, thus ZZ

Z bZ

d

f (x, y)dA =

f (x, y)dydx.

R

a

c

Warning: Some authors choose to write this as, Z

Z

b

d

dx a

f (x, y)dy. c

MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

21

By first accumulating across rows, then totaling their sums, we see that also, ¶ ZZ Z d µZ b f (x, y)dA = f (x, y)dx dy R

Z

c d

Z

=

a b

f (x, y)dxdy. c

a

We can choose to do the iterated integration in whichever order leads to the simplest calculations. Intuitively, the iterated integral, Z dZ b f (x, y)dxdy, c

a

may be thought of as representing our solid made up of cross-sectional laminae parallel to the x-z plane each of thickness dy. z

O

c

y

d

y x

Figure 16. A cross-sectional lamina parallel to the x-z plane. The lamina at a given y value has area Z b A(y) = f (x, y)dx a

and infinitesimal volume V (y) = A(y)dy. The volume of our solid is the ‘sum’ of these over all y between c and d; that is, ¶ Z d Z d µZ b A(y)dy = f (x, y)dx dy. c

c

a

This should be compared with the ‘slice method’ for finding volumes that you would have met either in high school or first year mathematics (MATH1120 or MATH1210). Example 4.1. Find the volume V of the solid half cylinder above the rectangle R = {(x, y) : −1 ≤ x ≤ 1, 0 ≤ y ≤ 2} in the x-y plane and below the surface z=

√

1 − x2 .

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MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

V =

ZZ √

1 − x2 dA

R 1 Z 2

Z = Z

−1 1

Z

−1 1

=

√

1 − x2 dydx

0

h √ iy=2 y 1 − x2 dx y=0

√

2 1 − x2 dx

= Z

−1 π/2

=

2 cos2 θ dθ,

−π/2

using the substitution x = sin θ

Z

π/2

=

1 + cos 2θ dθ −π/2

·

1 =π+ sin 2θ 2

¸π/2 −π/2

= π,

as expected!

Example 4.2. Evaluate, ZZ x cos xy dA, R

where R = {(x, y) : 0 ≤ x ≤ π/2, 0 ≤ y ≤ 1}. We can express this as the iterated integral, ZZ

Z

1

Z

π/2

x cos xy dA = R

however, to do the inner integral require integration by parts.

x cos xy dxdy, 0

R

0

x cos xy dx (with y regarded as a constant) would

MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

23

If instead, we reverse the order of integration, then the resulting integrations are easily performed; ZZ Z π/2 Z 1 x cos xy dA = x cos xy dydx R 0 0 ¶ Z π/2 µZ 1 = x cos xy dy dx, 0

0

x a constant for inner integral, Z π/2 = [sin xy]y=1 y=0 dx Z

0

π/2

=

sin x dx 0

= [− cos x]x=π/2 x=0 = 1. 4.3. Double Integrals Over General Regions. We now consider double integrals ZZ f (x, y)dA, D

where D is a general region in the plane (such as in figure 17). y

D

x

0

Figure 17. The region D. We will assume that D is a bounded region; that is, D is contained in some rectangle R = {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d}. (See figure 18). We subdivide the rectangle R into mn subrectanges Rij as before, and define, ZZ f (x, y)dA D X := lim f (xi , yj )∆A, m,n→∞

provided this limit exists.

{(i,j) : Rij ∩D6=∅}

24

MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA y

R D

x

0

Figure 18. The region D bounded by R. Equivalently,

ZZ

ZZ f (x, y)dA =

F (x, y)dA,

D

R

where,

( F (x, y) :=

f (x, y) if (x, y) ∈ D, 0 otherwise.

We will say that a plane region D is a type I region if it has the form; D = {(x, y) : g1 (x) ≤ y ≤ g2 (x), a ≤ x ≤ b}, where a and b are constants and g1 (x) and g2 (x) are continuous functions (see figure 19). y = g2(x)

y

y y = g2(x)

D

D

y = g1(x)

y = g1(x) 0

b

a

x

0

a

y y = g2(x)

y = g1(x) 0

a

b

x

Figure 19. Some type I regions.

b

x

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25

When D is a type I region we may, as before, think of ZZ f (x, y)dA D

as the ‘sum’ of an infinite number of limina parallel to the y-z plane each of thickness ‘dx’, but now the lamina at x will have area, Z

g2 (x)

A(x) =

f (x, y) dy. g1 (x)

See figure 20.

y d y = g2(x)

D y = g1(x)

c

0

x

a

b

x

Figure 20. A lamina at x, parallel to the y-z plane.

In this case we may therefore express our double integral as the iterated integral, ZZ

Z bZ

g2 (x)

f (x, y)dA = D

f (x, y) dydx. a

g1 (x)

2

Example 4.3. Find the volume V of the solid bounded above by the surface z = ex and below by the region D = {(x, y) : 0 ≤ y ≤ x, 0 ≤ x ≤ 1}, in the x-y plane (see figure 21).

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MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

z = e x2

z

1

1

y

1

x 0

2

Figure 21. The solid bounded above by the surface z = ex and below by the region D = {(x, y) : 0 ≤ y ≤ x, 0 ≤ x ≤ 1}

Z

1

Z

1

h

x

V = Z

0

= Z

2

ex dydx

0 2

yex

iy=x

0 1

y=0

dx

2

xex dx 0 · ¸x=1 1 x2 = (e − 1)/2. = e 2 x=0 =

Similarly, a type II region has the form, D = {(x, y) : h1 (y) ≤ x ≤ h2 (y), c ≤ y ≤ d}, where c and d are constants and h1 (y) and h2 (y) are continuous functions (see figure 22). y x = h2(y)

D x = h1(y)

0

x

Figure 22. A type II region.

MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

For such regions we have, ZZ

Z

d

Z

27

h2 (y)

f (x, y)dA =

f (x, y) dxdy.

D

c

h1 (y)

Some plane regions can be described as either a type I or a type II region. In these cases we are free to choose whichever order of integration leads to the iterated integral which is easiest to evaluate. Observe that we were able to evaluate ZZ 2 V := ex dA, D

where D = {(x, y) : 0 ≤ y ≤ x, 0 ≤ x ≤ 1}, by regarding D as a type I region and so express V as the iterated integral Z 1Z x 2 ex dydx. 0

0

If, however, we regard D as a type II region we are led to Z 1Z 1 2 V = ex dxdy, 0

an iterated integral involving the

R

y

2

ex dx which cannot be determined in closed form.

Example 4.4. Find the volume V of the solid between the x-y plane and the surface p z = x − y2, 0 ≤ x ≤ 4 We have, V =

ZZ p

x − y 2 dA,

D

where, described as a type I region, √ √ D = {(x, y) : − x ≤ y ≤ x, 0 ≤ x ≤ 4}. Thus, V may be expressed as the iterated integral, Z 4 Z √x p V = x − y 2 dydx. √ 0

− x

Attempts to evaluate this lead to extremely intricate calculations. Try it! If, however, we describe D as a type II region, D = {(x, y) : y 2 ≤ x ≤ 4, −2 ≤ y ≤ 2}, we are led to,

Z

2

Z

4

p

x − y 2 dxdy.

V = −2

y2

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Thus,

Z

2

Z

4

p

x − y 2 dxdy

V = Z

−2 2

= Z

−2

·

y2

2 (x − y 2 )3/2 3

¸x=4 dy x=y 2

2

2 (4 − y 2 )3/2 dy 3 −2 Z 32 π/2 = cos4 θ dθ, 3 −π/2 using the substitution y = 2 sin θ = 4π, =

as expected (why?) 4.4. Properties of Double Integrals. Here are some useful properties of double integrals that follow directly from the definition. ZZ f (x, y) + g(x, y) dA (1) ZDZ ZZ = f (x, y) dA + g(x, y) dA. ZZ D ZZ D (2) cf (x, y) dA = c f (x, y) dA, D

D

where c is a constant. (3) If D = D1 ∪ D2 , where D1 and D2 are disjoint regions of the plane (that is, D1 ∩ D2 = ∅), then ZZ ZZ ZZ f (x, y) dA = f (x, y) dA + f (x, y) dA. D

D1

D2

This may be used to evaluate double integrals over regions that are neither of type I or of type II, but that can be built from such regions (see figure 23). y

y

D2 D1

D

0

x

0

x

Figure 23. D is nether type I nor type II but D1 is type I and D2 is type II.

MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

29

ZZ (4)

1 dA = A(D), D

where A(D) is the area of the plane region D. (5) If f (x, y) ≤ g(x, y), for all (x, y) in D, then ZZ ZZ f (x, y) dA ≤ g(x, y) dA. D

D

(6) Consequently, if m ≤ f (x, y) ≤ M , for all (x, y) in D, where m and M are constants, then ZZ m A(D) ≤ f (x, y) dA ≤ M A(D). D

4.5. Double Integrals In Polar Coordinates. Some functions of two variables and some plane regions are more simply expressed using polar coordinates (r, θ), in which case it is often expedient to express double integrals involving these functions and regions in terms of the variables r and θ rather than x and y. 2 2 2 For example; f (x, y) = e−(x +y ) = e−r and regions such as those shown in figure 24,

y x2 + y2 = 1

y x2 + y2 = 4

R 0

R x

0

x x2 + y2 = 1

Figure 24. R = {(r, θ)|0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π} and R = {(r, θ)|1 ≤ r ≤ 2, 0 ≤ θ ≤ π}. We can achieve this writing, ZZ f (x, y)dA := D X lim m,n→∞

f (ri∗ cos θj∗ , ri∗ sin θj∗ ) A(Rij ),

{(i,j) : Rij ∩D6=∅}

where Rij are now the ‘polar rectangles’Rij := {(r, θ) : ri−1 ≤ r ≤ ri , θj−1 ≤ θ ≤ θj } (see figure 25).

30

MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA θ = θj θ = θj _ 1 (r*i ,θ*j )

Rij

∆θj r = ri _ 1

r = ri

0

Figure 25. Polar rectangles. and A(Rij ) is the area of Rij : 1 1 A(Rij ) = (r + ∆r)2 ∆θ − r2 ∆θ 2 2 ≈ r∆r∆θ. Thus, for a region D of the form shown in figure 26,

r = h2(θ)

θ=β

D

θ=α r = h1(θ)

β α 0

Figure 26. R = {(r, θ)|h1 (θ) ≤ r ≤ h2 (θ), α ≤ θ ≤ β}.

we have, as an iterated integral in terms of polar coordinates, ZZ

Z

β

Z

h2 (θ)

f (x, y) dA = D

f (r cos θ, r sin θ) rdrdθ. α

h1 (θ)

NOTE: In terms of polar coordinates the ‘infinitesimal’ area dA = rdrdθ. See figure 27.

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31

dA dθ dr r dθ r

0

Figure 27. dA = rdrdθ.

Example 4.5. For D := {(x, y) : x2 + y 2 ≤ 4} we have, ZZ

Z 2

2

2π

Z

Z

0 2π · 2

(1 + x + y ) dA = D

Z

2π

0

Z

2

=

3

(r + r )drdθ = Z

0

0

0

2

(1 + r2 )rdrdθ r r4 + 2 4

¸r=2 dθ r=0

2π

=

6dθ = 12π. 0

Example 4.6. The volume of the hemisphere H := {(x, y, z) : x2 + y 2 + z 2 ≤ a2 , z ≥ 0}, is, ZZ z dA, D

where the disk {(x, y) : x2 + y 2 ≤ a2 }, Z 2πD Z is a√ a2 − r2 rdrdθ = 0

0

= ··· 2 = πa3 , as expected. 3 ZZ

Z

Example 4.7.

π/4

Z

2 sin θ

x dA = D

0

0

1 r cos θ rdrdθ = · · · = , 6

where D is the region bounded above by the line y = x and below by the circle x2 + y 2 − 2y = 0.

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MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

4.6. Change of variables in double integrals. We have seen that some double integrals are more conveniently evaluated when expressed in terms of the polar coordinate variables r and θ rather than the Cartesian variables x and y. We wish to explore this situation more generally by investigating how to express the double integral ZZ f (x, y) dxdy D

in terms of new variables u and v, where x = x(u, v) and y = y(u, v). Example 4.8. (1) x = x(u, v) := u2 − v 2 and y = y(u, v) := 2uv. (2) x = x(r, θ) := r cos θ and y = y(r, θ) := r sin θ, representing a change to polar coordinates. It will be usefull to think of the relations x = x(u, v) and y = y(u, v). as defining a transformation T of the u-v plane to the x-y plane, which transforms (maps) the point (u, v) in the u-v plane to the point (x, y) = T (u, v) := (x(u, v), y(u, v)) in the x-y plane. Example 4.9. (1) Under the transformation of example (1) above; (x, y) = T (u, v) := (u2 −v 2 , 2uv) the point (1, 1) in the u-v plane is transformed to the point (0, 2) in the x-y plane. (2) Similarly, (x, y) = T√ (r, θ) √ := (r cos θ, r sin θ), maps the point (2, π/4) in the r-θ plane to the point ( 2, 2), the point in the x-y plane whose polar coordinates are (2, π/4). For technical reasons we will assume that both the functions x(u, v) and y(u, v) have continuous partial derivatives and that the transformation T they define is one-toone; that is, distinct points in the u-v plane are mapped by T to distinct points in the x-y plane, so the inverse transformation T −1 exists. This last requirement often forces us to impose restrictions on the domain of T (the admissible values of u and v). In the case of polar coordinates, a restriction to principal values, while for the transformation of example (1) above; T (u, v) := (u2 − v 2 , 2uv) requiring, say u to be positive (clearly the points (−u, −v) and (u, v) share the same image under the transformation T ).

MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

33

NOTE: The domain u ≥ 0 is partitioned into regions where T (u, v) has x ≤ 0, y ≥ 0, x ≥ 0, y ≤ 0,

x ≥ 0, y ≥ 0, x ≤ 0, y ≤ 0. u x< 0 y>0 x>0 y>0 x>0 y < 0 x y

v

<0 <0

Figure 28. The domain of T (u, v) := (u2 − v 2 , 2uv).

Under T a region R in the u-v plane is transformed into a region, D = T (R) := {T (u, v) = (x(u, v), y(u, v)) : (u, v) ∈ R}, in the x-y plane. Example 4.10. (1) Under the transformation of example (1) above; (x, y) = T (u, v) := (u2 − v 2 , 2uv), we see that the square R = [0, 1] × [0, 1] = {(u, v) : 0 ≤ u ≤ 1, 0 ≤ v ≤ 1} is transformed onto the region D bounded by the parabolas; x = ±(1 − y 2 /4) and the x - axis. (2) Similarly, T (r, θ) := (r cos θ, r sin θ), transforms the rectangle, R = {(r, θ) : 1 ≤ r ≤ 2, 0 ≤ θ ≤ π/4}, in the r-θ plane to the region D in the x-y plane illustrated below. These examples illustrate how a suitable transformation T can associate a relatively simple region R in the u-v plane (in both cases a rectangle) with a more complicated region D = T (R) in the x-y plane. Our goal is to express ZZ f (x, y) dxdy D

34

MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

as a double integral of the function F (u, v) := f (x(u, v), y(u, v)) over the (simpler) region R = T −1 (D) in the Zu-v Z plane. We do this by realizing

f (x, y) dxdy as follows. D

As before, in the u-v plane we subdivide a rectangle containing the region R into mn subrectangles Rij , with centre the point (ui , vj ), and each of area A(Rij ) = ∆u∆v. This gives us a covering of D = T (R) by transformed rectangles T (Rij ) and we have, ZZ X f (x(ui , vj ), y(ui , vj ))A(T (Rij )) f (x, y) dxdy = lim D

m,n→∞

{i, j : (ui , vj ) ∈ R}

It remains to determine how the area A(T (Rij )), of the transformed subrectangle is related to the area ∆u∆v of Rij . NOTE: Observe that this is precisely how we developed the double integral for polar coordinates from first principles; covering D with ‘polar rectangles’ the image under the polar transform T of Rij = [ri , ri + ∆r] × [θj , θj + ∆θ]. There we noted A(T (Rij )) = rA(Rij ) = r∆r∆θ. If we let (u0 , v0 ) denote the bottom lefthand corner of the subrectangle Rij and r(u, v) := T (u, v) = (x(u, v), y(u, v)) be the position vector of the image in the xy plane of the point (u, v) we see that for ∆u and ∆v small, A(T (Rij )) is approximately the area of the parallelogram with sides r(u0 + ∆u, v0 ) − r(u0 , v0 ) and r(u0 , v0 + ∆v) − r(u0 , v0 ). Further, for ∆u and ∆v small r(u0 + ∆u, v0 ) − r(u0 , v0 ) ≈ ru (u0 , v0 )∆u and r(u0 , v0 + ∆v) − r(u0 , v0 ) ≈ rv (u0 , v0 )∆v. Thus, A(T (Rij )) is approximately the area of the parallelogram with sides ru (u0 , v0 )∆u and rv (u0 , v0 )∆v.

MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

35

Thinking of these as vectors in R3 , we know (MATH1110, or MATH1210, or Stewart 13.4) that this area is the length (magnitude) of their cross product, so A(T (Rij )) ≈ k(ru (u0 , v0 )∆u) × (rv (u0 , v0 )∆v)k = kru (u0 , v0 ) × rv (u0 , v0 )k∆u∆v ¯° °¯ °¯i j k¯¯ ° ° ° ¯ ∂x ∂y ¯ ¯ ° ∆u∆v 0 =° ∂u ¯° ° ¯ ∂u ° ¯ ∂x ∂y 0 ¯ ° ∂v ∂v ¯ ¯ ∂x ∂y ¯ ¯ ¯¯ ¯¯ ∂u ¯ ¯ ∆u∆v = ¯¯ ¯¯ ∂u ∂y ¯ ¯ ∂x ∂v

∂v

The determinant appearing in this last expression is known as the Jacobian of the transformation T (u, v) = (x(u, v), y(u, v)) and is denoted by ¯ ∂x ∂y ¯ ¯ ¯ ∂(x, y) ∂u ¯ := ¯¯ ∂u ∂y ¯ ∂x ∂(u, v) ∂v

∂v

∂x ∂y ∂x ∂y = − . ∂u ∂v ∂v ∂u For example, for the transformation T (u, v) := (u2 − v 2 , 2uv) we have ¯ ¯ ∂(x, y) ¯¯2u −2v ¯¯ = ∂(u, v) ¯2v 2u ¯ = 4u2 + 4v 2 . Substituting the approximation,

¯ ¯ ¯ ∂(x, y) ¯ ¯ ∆u∆v, A(T (Rij )) ≈ ¯¯ ∂(u, v) ¯

into our previous expression for the double integral and taking the limit as m, n → ∞ (∆u, ∆v → 0) leads to, ZZ f (x, y) dxdy D=T (R)

ZZ =

¯ ¯ ¯ ∂(x, y) ¯ ¯ ¯ dudv. f (x(u, v), y(u, v)) ¯ ¯ ∂(u, v) R

A result known as the change of variable formula for double integrals, which should be compared to the familiar change of variable formula Z x(b) Z b dx f (x) dx = f (x(u)) du. du x(a) a for functions of a single variable. As an example, for the double integral of any function f (x, y) over the region D bounded by the parabolas x = ±(1 − y 2 /4) and the x - axis we have, using the transformation

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MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

T (u, v) := (u2 − v 2 , 2uv), that ZZ f (x, y) dxdy D Z 1Z 1 = f (u2 − v 2 , 2uv)(4u2 + 4v 2 ) dudv. 0

0

Thus, for instance, ZZ y dxdy Z 1Z 1 = 2uv(4u2 + 4v 2 ) dudv 0 0 Z 1Z 1 = 8(vu3 + v 3 u) dudv D

Z

0

0 1

= Z

0 1

= 0

·

vu4 v 3 u2 8 + 4 2

¸u=1 dv u=0

£ ¤1 (2v + 4v 3 )dv = v 2 + v 4 0 = 2.

In the case of polar coordinates, the transformation (x, y) = T (r, θ) = (r cos θ, r sin θ), has Jacobian, ¯ ¯ ∂(x, y) ¯¯ cos θ sin θ ¯¯ =¯ = r, −r sin θ r cos θ¯ ∂(r, θ) confirming our previous result; ZZ ZZ f (x, y) dxdy = f (r cos θ, r sin θ) r drdθ. D

R

∂(x, y) of the transformation T represents the factor by which ∂(u, v) an infinitessimal area dudv in the u-v plane is scaled when it is mapped by T to the x-y plane. Thus, we see that the Jacobian of T −1 , Note: The Jacobian,

· ¸−1 ∂(x, y) ∂(u, v) = . ∂(x, y) ∂(u, v) So far we have viewed double integrals as representing the volume of a solid, however, double integrals have many interpretations, and find numerous applications throughout mathematics and in the physical, economic and engineering sciences (for example, see Stewart, sections 16.5 and 16.6). We consider just one of these.

MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

37

4.7. Area of a surface as a double integral. We consider the problem of finding the area A(S) of the patch S of the surace z = f (x, y) above a (bounded) region D in the x-y plane. In the now familiar way, we subdivide a rectangle containing D into mn subrectangles Rij and take the point (xi , yj ) to be the centre of Rij . The tangent plane to the surface at (xi , yj ) provides an approximation to the surface near this point. The area ∆Aij of the rectangular patch of the tangent plane above Rij therefore gives us an approximation for the area of the surface above this subretangle, and X ∆Aij , A(S) := lim m,n→∞

{(i,j) : (xi ,yj )∈D}

provided the limit exists. Now, if θ is the acute angle a normal to the tangent plane at (xi , yj ) makes to the vertical, then we see that ∆Aij cos θ is the area of the subrectangle Rij , so, ∆Aij =

∆x∆y . cos θ

Further, thinking of our surface as the level surface F (x, y, z) := z − f (x, y) = 0, the direction of a normal to the tangent plane at the point on the surface above (xi , yj ) is, µ ¶ ∂f ∂f n = ∇F = − , − , 1 , ∂x ∂y evaluated at (xi , yj ), so cos θ =

1 1 n · (0, 0, 1) = r . knk ¡ ∂f ¢2 ³ ∂f ´2 + ∂y + 1 ∂x

Thus,

s ∆Aij =

µ 1+

∂f ∂x

¶2

µ +

∂f ∂y

¶2 ∆x∆y.

[For an alternative derivation of this, see Stewart, section 16.6.] And so, taking limits in our expression for A(S) leads to;   s µ ¶2 µ ¶2     ∂f ∂f    A(S) =  1+ + dxdy.     ∂x ∂y D

p Example 4.11. The surface area A of the cone z = k x2 + y 2 lying between the planes z = 0 and z = h > 0. This section of the conical surface lies above the disk

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MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

D = {(x, y) : x2 + y 2 ≤ (h/k)2 }, so ZZ r ³ ´2 ³ ´2 p p A= 1 + kx/ x2 + y 2 + ky/ x2 + y 2 dxdy D ZZ √ 2 = 1+k 1 dxdy D √ √ = 1 + k 2 A(D) = 1 + k 2 π (h/k)2 √ = πr r2 + h2 , where r = h/k is the radius of the cone’s base.

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39

5. Triple Integrals 5.1. Triple Integrals Over General Regions. We have defined single integrals for functions of one variable, and double integrals for functions of two variables. Now we want to define triple integrals for functions of three variables. These are useful in computing the mass and centre of mass for solid objects of varying densities, the moments of inertia of rotating bodies, the total electrical charge on non-conducting bodies, and in many other physical applications. As for double integrals, we begin with the simplest case of a function defined on a three dimensional rectangular box B = [a, b] × [c, d] × [r, s] = {(x, y, z) : a ≤ x ≤ b, c ≤ y ≤ d, r ≤ z ≤ s}. We begin as always by dividing our box B up into small sub-boxes: • divide [a, b] up into l subintervals [xi−1 , xi ], each of length ∆x = (b−a) ; l d−c • divide [c, d] up into m subintervals [yi−1 , yi ] of length ∆y = m ; and • divide [r, s] up into n subintervals [zi−1 , zi ] of length ∆z = s−r . n The net effect is to divide B into lmn sub-boxes: Bi,j,k = [xi−1 , xi ] × [yj−1 , yj ] × [zk−1 , zk ], each with volume ∆V = ∆x∆y∆z =

(b − a)(d − c)(s − r) . lmn

∆z z

z

∆x ∆y

B

y

y x

x

Figure 29. The volume element ∆V . We approximate functions f : B −→ R by simpler functions which are constant on each Bi,j,k . So if f (x, y, z) is a function of three variables whose domain contains B, we define the triple Riemann sum Sl,m,n :=

l X m X n X

∗ ∗ )∆V , zi,j,k f (x∗i,j,k , yi,j,k

i=1 j=1 k=1 ∗ ∗ where (x∗i,j,k , yi,j,k , zi,j,k ) is the centre of the sub-box Bi,j,k .

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To visualise what is going on, imagine that f describes the local density (in kg/m3 ) of a B-shaped brick. Then Sl,m,n approximates the mass of the box B by assuming the density is constant within each sub-box. By analogy with the single- and double-integrals defined previously, we define the triple integral ZZZ f (x, y, z)dV := lim Sl,m,n , l,m,n−→∞

B

provided this limit exists. RRR Thinking of f as the density function again, f (x, y, z)dV tells us the mass of B RRR B. Provided that f is continuous, f (x, y, z)dV always exists, and is the same no B ∗ ∗ ∗ matter which sample point (xi,j,k , yi,j,k , zi,j,k ) we chose in each Bi,j,k . Just as with double integrals, calculating triple Riemann sums and taking the limit is generally impractical. Instead, we evaluate triple integrals as iterated integrals: ZZZ

Z b³Z d³Z

´ ´ f (x, y, z)dz dy dx

s

f (x, y, z)dV = B

a

c

r

As with iterated integrals in two variables, we integrate from the inside to the outside of the brackets, keeping x and y fixed (that is, treating them as constants) while we integrate with respect ot z, and then keeping x fixed while we integrate with respect to y. Also, as with double integrals, we generally drop the parentheses and write Z bZ dZ

s

f (x, y, z) dz dy dx a

c

r

Here we integrate with respect to z first, then RRR y, then x. However, we may choose any order of integration. That is to say, while f (x, y, z)dV can be expressed as B ZZZ

Z bZ dZ

s

f (x, y, z)dV = B

f (x, y, z)dzdydx a

c

r

as above, it is equally valid to evaluate it as ZZZ

Z sZ bZ

d

f (x, y, z)dV = B

f (x, y, z)dydxdz, r

a

c

or any of the the other four permutations of x, y and z. . Example 5.1. Evaluate

RRR B

6xzeyz dV where B = [0, 1] × [0, 3] × [1, 2].

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41

Solution: We notice that integrating with respect to y first will eliminate the factor of z: ZZZ Z 1Z 2Z 3 yz 6xze dV = 6xzeyz dy dz dx B 0 1 0 Z 1Z 2 h iy=3 = 6xeyz dz dx y=0 0 1 Z 1Z 2 = 6x(e3z − 1) dz dx 0 1 Z 1h iz=2 3z = 2x(e − 3z) dx z=1 0 Z 1 = (e6 − e3 − 3)2x dx 0 h i1 = (e6 − e3 − 3)x2 0

6

3

= e − e − 3. Now we want to deal with regions that are not rectangular prisms (boxes). Technically, to define the triple integral over a bounded region E, we first choose a box B that contains E. Given a function f defined on E, we define a function F on B by setting F (x, y, z) := f (x, y, z) for (x, y, z) ∈ E and setting F (x, y, z) = 0 for (x, y, z) ∈ B \ E. Now we define ZZZ ZZZ f (x, y, z) dV := E

F (x, y, z) dV. B

This definition of the triple integral satisfies all the properties we would expect it to satisfy: RRR RRR RRR • RRRE (f + g)(x, y, z) dV = RRRE f (x, y, z) dV + g(x, y, z) dV E • k · f (x, y, z) dV = k · f (x, y, z) dV E E RRR RRR • if f (x, y, z) ≤ g(x, y, z) for all (x, y, z) ∈ E, then f (x, y, z) dV ≤ g(x, y, z) dV . E E • if E1 ∪ E2 = E and E1 ∩ E2 = ∅, then ZZZ ZZZ ZZZ f (x, y, z) dV = f (x, y, z) dV + f (x, y, z) dV E

E1

E2

However, this is not a very practical way to evaluate triple integrals. In practice, as for double integrals, we will consider simple regions E = {(x, y, z) : a ≤ x ≤ b, u1 (x) ≤ y ≤ u2 (x), v1 (x, y) ≤ z ≤ v2 (x, y)} where u1 , u2 , v1 and v2 are continuous functions. These are the three-dimensional versions of the type I regions we saw for double integrals.

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MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA z z = v2(x, y)

E z = v1(x, y) O a

y

y = u1(x)

y = u2(x)

D

b x

Figure 30. A type I region. Writing D for the “shadow” of E in the xy-plane, we define ZZZ Z Z ³Z v2 (x,y) ´ f (x, y, z)dV = f (x, y, z) dz dA E

D

Z bZ

v1 (x,y)

u2 (x)Z v2 (x,y)

f (x, y, z) dz dy dx.

= a

u1 (x)

v1 (x,y)

Of course, we can handle regions described in the same way but with the variables permuted simply by changing the order of integration. For example, if E = {(x, y, z) : a ≤ y ≤ b, u1 (y) ≤ z ≤ u2 (y), v1 (y, z) ≤ x ≤ v2 (y, z)}, then

ZZZ

Z bZ

u2 (y)Z v2 (y,z)

f (x, y, z)dV = E

f (x, y, z) dx dz dy. a

u1 (y)

v1 (y,z)

Example 5.2. Integrate the function f (x, y, z) = xyz over the region bounded by: • the paraboloid z = x2 + y 2 , • the plane z = 1, • the xz-plane (y = 0), and • the yz-plane (x = 0). Solution: We express the region as E = {x, y, z : 0 ≤ z ≤ 1, 0 ≤ y ≤ p 0 ≤ x ≤ (z − y 2 )}.

√

z,

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z

y x Figure 31. The region bounded by z = x2 + y 2 , z = 1, y = 0 and x = 0

Now we can calculate ZZZ f (x, y, z)dV E

Z 1Z

√

√

zZ

z−y 2

= 0

Z 1Z = 0

= = = = =

0

√

0

z

h1

2 0 √ 1Z z

x2 yz

xyz dx dy dz i √ 2 x=

x=0

z−y

dy dz

Z 1 (z − y 2 )yz dy dz 2 0 0 √ Z 1 1 h 1 2 2 1 4 iy= z y z − y z dz 2 0 2 4 y=0 Z 1 1 3 z dz 8 0 1 h 1 4 i1 z 8 4 0 1 . 32

5.2. Change of Variables in Triple Integrals. As in single and double integrals, making a variable substitution can make triple integrals much simpler. R often 3 For example, when integrating x (1−1/x4 )dx, it is easiest to proceed by making the substitution u = x4 before proceeding. To make this work, though, we have to replace dx with 4x3 dx. We saw that to achieve the same result in double integrals, one must use the Jacobian: if x = g(u, v), y = h(u, v) maps a region C in uv-space to the region D in xy-space,

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then we can change variables in the double integral: ¯ ¯ ZZ ZZ ¯ ∂(x, y) ¯ ¯ ¯ du dv f (x, y)dA = f (g(u, v), h(u, v)) ¯ ¯ ∂(u, v) D C A similar procedure works for triple integrals. If T is a transformation which takes a region R in uvw-space to the region S in xyz-space via the maps x = x(u, v, w) y = y(u, v, w) z = z(u, v, w), then the Jacobian of T is defined to be the determinant  ∂x ∂x ∂x    ∂(x, y, z) := det   ∂(u, v, w) 

∂u

∂v

∂w

∂y ∂u

∂y ∂v

∂y ∂w

∂z ∂u

∂z ∂v

∂z ∂w

    

Theorem 5.3. Suppose that T (u, v, w) := (x(u, v, w), y(u, v, w), z(u, v, w)) is a one-to-one and once differentiable map which takes a region R in uvw-space to a region S in xyz-space. Suppose that f is continuous on S, and that both R and S are regions of the type described above. ¯ ¯ ZZZ ZZZ ¯ ∂(x, y, z) ¯ ¯ du dv dw f (x, y, z) dV = f (T (u, v, w)) ¯¯ ∂(u, v, w) ¯ S R 5.3. Triple Integrals in Cylindrical and Spherical Coordinates. We apply the three-variable Jacobian to integration over solid regions in cylindrical and spherical coordinates. 5.3.1. Cylindrical Coordinates. Consider the transformation from cylindrical coordinates to cartesian coordinates: x = r cos(θ) y = r sin(θ) z = z The Jacobian is given by



  ∂(x, y, z) = det   ∂(r, θ, z)  

∂x ∂r

∂x ∂θ

∂x ∂z

∂y ∂r

∂y ∂θ

∂y ∂z

∂z ∂r

∂z ∂θ

∂z ∂z

     

 cos(θ) −r sin(θ) 0 = det  sin(θ) r cos(θ) 0  0 0 1 = r cos2 (θ) + r sin2 (θ) = r.

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So, in terms of cylindrical coordinates

ZZZ

ZZZ f (x, y, z) dxdydz =

f (r cos θ, r sin θ, z) rdrdθdz,

D

S

where D = T (S).

z

dz

dθ r

dr r dθ

Figure 32. A volume element in cylindrical coordinates.

Example 5.4. Calculate the integral of f (x, y, z) := x2 + y 2 over the region E bounded 1 by the planes z = 1 and z = 2 and the hyperboloid z 2 = x2 +y 2. Solution: Changing to cylindrical coordinates, we want the integral of the function

f (r cos(θ), r sin(θ), z) = r2 cos2 (θ) + r2 sin2 (θ) = r2

over the region bounded by z = 1 and z = 2 and the surface z 2 = 1/r2 , ie. r = 1/z. Thus

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MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

ZZZ

Z 2

2

2π

Z

2π

Z

2π

Z

2

Z

2

Z

1/z

x + y dV = E

Z

0

1

0 1/z

= Z

0

= 0

= = = = =

1

¯ ¯ ¯ ∂(x, y, z) ¯ ¯ ¯ dr dz dθ r ¯ ∂(r, θ, z) ¯ 2

r3 dr dz dθ

0

2

h r4 i1/z

4 1 2π Z 2

0

dz dθ

Z 1 z −4 dz dθ 4 0 1 Z 2π h 1 −1 −3 i2 z dθ 4 0 3 1 Z 1 2π 7 dθ 4 0 24 7 h i2π θ 96 0 7π . 48

5.3.2. Spherical Coordinates. Consider the transformation from spherical coordinates to cartesian coordinates:

x = ρ sin(φ) cos(θ) y = ρ sin(φ) sin(θ) z = ρ cos(φ)

The Jacobian is given by    ∂(x, y, z) = det  J=  ∂(ρ, θ, φ) 

We can calculate it as follows:

∂x ∂ρ

∂x ∂θ

∂x ∂φ

∂y ∂ρ

∂y ∂θ

∂y ∂φ

∂z ∂ρ

∂z ∂θ

∂z ∂φ

   .  

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47

¯ ¯ ¯ sin φ cos θ −ρ sin φ sin θ ρ cos φ cos θ ¯ ¯ ¯ J = ¯¯ sin φ sin θ ρ sin φ cos θ ρ cos φ sin θ ¯¯ ¯ cos φ 0 −ρ sin φ ¯ ¯ ¯ ¯ −ρ sin φ sin θ ρ cos φ cos θ ¯ ¯ = cos φ ¯¯ ρ sin φ cos θ ρ cos φ sin θ ¯ ¯ ¯ ¯ sin φ cos θ −ρ sin φ sin θ ¯ ¯ + (−ρ sin φ) ¯¯ sin φ sin θ ρ sin φ cos θ ¯ = −ρ2 cos2 φ sin φ(sin2 θ + cos2 θ) − ρ2 sin3 φ(cos2 θ + sin2 θ) = −ρ2 sin φ(cos2 φ + sin2 φ) = −ρ2 sin φ. Since 0 ≤ φ ≤ π in spherical coordinates, we always have sin φ ≥ 0, and hence ¯ ¯ ¯ ∂(x, y, z) ¯ 2 ¯ ¯ ¯ ∂(ρ, θ, φ) ¯ = ρ sin φ. So, in terms of spherical coordinates ZZZ f (x, y, z) dxdydz = D ZZZ f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) ρ2 sin φdρdφdθ, S

where again D = T (S). z ρ sinφ dθ

φ

dρ

ρ

O

ρdφ dθ

y

x

Figure 33. A volume element in spherical coordinates.

Example 5.5. Integrate the function f (x, y, z) := (x2 + y 2 + z 2 )−1 over the region E above the xy-plane and between the spheres of radii 1 and 2 about the origin.

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Solution: Changing to spherical coordinates gives f (ρ sin(φ) cos(θ), ρ sin(φ) sin(θ), ρ cos(φ)) 1 = 2 2 2 ρ (sin φ cos θ + sin2 φ sin2 θ + cos2 φ) 1 = 2 2 ρ (sin φ + cos2 φ) 1 = 2, ρ and E is simply the region 1 ≤ ρ ≤ 2, 0 ≤ φ ≤ π2 . Hence ZZZ

Z 2

2

π/2Z 2πZ 2

2 −1

(x + y + z ) dV = E

Z

0

0 1 π/2Z 2πZ 2

=

1 ρ2

¯ ¯ ¯ ∂(x, y, z) ¯ ¯ ¯ ¯ ∂(ρ, θ, φ) ¯ dρ dθ dφ

sin(φ) dρ dθ dφ Z

0

0

π/2Z 2π

Z

0

π/2Z

Z

0

Z

0

=

1

h

iρ=2 sin(φ)ρ dθ dφ ρ=1

0 2π

=

sin(φ) dθ dφ 0

π/2

=

h

sin(φ)θ

iθ=2π θ=0

π/2

=

2π sin(φ) dφ 0

h iπ/2 = 2π − cos(φ) 0

= 2π.

dφ

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6. Line and Surface Integrals 6.1. Line integrals. Let f (x, y) be a function of two variables defined on some domain D and let C be a curve lying entirely in D given parametrically by, x = x(t),

y = y(t),

a ≤ t ≤ b,

r(t) := x(t)i + y(t)j,

a ≤ t ≤ b.

or equivalently in vector form, We further assume that C is a smooth curve; that is, r0 (t) exists, is continuous and non-zero for all t in the interval [a, b]. We divide the interval [a, b] into n subintervals of equal length, [ti , ti+1 ], where t0 = a, ti+1 = ti + ∆t, for i = 0, 1, 2, · · · n − 1 and ∆t = (b − a)/n, so tn = b (see figure 34) and define the line integral of f along C with respect to arc length by Z n−1 X f (x, y) ds := lim f (x(ti ), y(ti ))∆si , n→∞

C

i=0

provided this limit exists, where ∆si is the length of the subarc of C joining the points (x(ti ), y(ti )) and (x(ti+1 ), y(ti+1 )).

y

(x(a), y(a))

(x(ti-1), y(ti-1))

(x(ti+1), y(ti+1))

(x(ti), y(ti))

C (x(b), y(b))

0

x

Figure 34. The intervals induced on C by partitioning [a,b] into equal subintervals. Thus, writing ∆x(ti ) := x(ti + ∆t) − x(ti ) and ∆y(ti ) := y(ti + ∆t) − y(ti ) we have, p ∆si ≈ ∆x(ti )2 + ∆y(ti )2 sµ ¶2 µ ¶2 ∆y(ti ) ∆x(ti ) + ∆t = ∆t ∆t sµ ¶ µ ¶2 2 dx dy ≈ + ∆t, dt dt provided ∆t is small enough (that is, n is large enough), where the derivatives are evaluated at t = ti .

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Taking the limit as n → ∞ we therefore have,

Z

Z f (x, y) ds = C

sµ

b

f (x(t), y(t)) Z

a b

=

dx dt

¶2

µ +

dy dt

¶2 dt

f (x(t), y(t)) kr0 (t)k dt,

a

whenever the righthand integral exists. Intuitively we may interpret this line integral as the area of a ‘fence’ with base the curve C in the x-y plane and whose height above the point (x, y) on C is f (x, y) (see figure 35).

z

}

0

C

y f (x, y)

(x, y)

x

Figure 35. A line integral interpreted as the area of a ‘fence’˙

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We can extend the definition to include the case when C is a piecewise-smooth curve; that is, C is the union of a finite number of contiguous smooth curves C1 , C2 , · · · , Ck , by writing Z Z Z f (x, y) ds := f (x, y) ds + · · · + f (x, y) ds. C

C1

Ck

Example 6.1. Suppose C = C1 ∪ C2 , where C1 consists of the line segment from (0, 0) to (1, 0) and C2 is the semicircle given by x = cos t, y = sin t with 0 ≤ t ≤ π. Then, Z Z Z 2 2 1 + x y ds = 1 + x y ds + 1 + x2 y ds C

Z

C1 1

Z

C2

p 1dx + (1 + cos2 t sin t) sin2 t + cos2 t dt 0 0 · ¸π cos3 t =1+ t− 3 0 5 = + π. 3 π

=

We can extend line integrals with respect to arc length to curves in R3 . If a curve C in R3 is given by, r(t) = x(t)i + y(t)j + z(t)k, for a ≤ t ≤ b, then

Z

Z

b

f (x, y, z) ds := C

f (r(t))kr0 (t)k dt.

a

6.2. Surface Integrals. Let S be a surface in R3 given parametrically by r(u, v) = (x(u, v), y(u, v), z(u, v)), for (u, v) ranging over some bounded domain D in the u-v plane, and for which the component functions x(u, v), y(u, v) and z(u, v) have continuous partial derivatives. Let R = [a, b] × [c, d] be a rectangle in the u-v plane containing D. For each pair of natural numbers m and n, let ∆u = (b − a)/m and ∆v = (d − c)/n and divide R into the mn subrectangles Rij = [ui , ui+1 ] × [vj , vj+1 ], where ui = a + i∆u and vj = c + j∆v, for i = 0, 1, · · · , m − 1 and j = 0, 1, · · · , n − 1. If Rij is contained in D, we denote by ∆Sij the area of the patch of surface Sij = {r(u, v) : (u, v) ∈ Rij } , See figure 36. Thus, Sij is bounded by the curves: r(u, vj ),

ui ≤ u ≤ ui+1 ,

r(ui , v),

vj ≤ v ≤ vj+1 ,

r(u, vj+1 ),

ui ≤ u ≤ ui+1

r(ui+1 , v),

vj ≤ v ≤ vj+1

and

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MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA z

v

D

(u, v)

Rij

r (u, v)

Sij

0 y

0

u

x

Figure 36. The elements Rij and Sij . By arguments similar to those used in developing line integrals, for m and n sufficiently large ∆Sij is approximately the area of the parallelogram with adjacent sides ru (ui , vj )∆u and rv (ui , vj )∆v. . That is, ∆Sij ≈ kru × rv k∆u∆v, where the derivatives in the right hand side are evaluated at (ui , vj ). If f (x, y, z) is a scalar valued function whose domain contains the surface S, then the surface integral ZZ X f dS := lim f (r(ui , vj ))∆Sij m,n→∞

S

ZZ =

{(i,j):Rij ⊆D}

f (r(u, v))kru × rv k dudv, D

provided this limit (equivalently, the right hand integral) exists. Example 6.2. Find

ZZ z dS, S

where S is the hemisphere with centre the origin and radius 1 sitting above the x-y plane. Using spherical coordinates, S may be described parametrically by, r(θ, φ) = (cos θ sin φ, sin θ sin φ, cos φ), for 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π/2. Thus, ZZ Z z dS = S

0

2π

Z

π/2

cos φkrθ × rφ k dφdθ. 0

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53

Now, rθ × rφ ¯ ¯ ¯ ¯ i j k ¯ ¯ 0 ¯¯ = ¯¯− sin θ sin φ cos θ sin φ ¯ cos θ cos φ sin θ cos φ − sin φ¯ = −(cos θ sin2 φ, sin θ sin2 φ, sin φ cos φ) Consequently, krθ × rφ k = sin φ and

ZZ

Z

2π

Z

π/2

z dS = S

cos φ sin φ dφdθ = π. 0

0

REMARKS: 1. The area of a surface S in R3 is given by, ZZ 1 dS. S

2. If the surface S is specified by z = f (x, y), for (x, y) ranging over some domain D, then we can parameterize it by r(x, y) = (x, y, f (x, y)) for (x, y) ∈ D. In this case,

°¯ ° ¯i j °¯ ¯ krx × ry k = ° ° ¯1 0 ° ¯0 1 s =

So, ZZ

µ 1+

¯° k ¯¯ ° ∂f ¯ ° ° ∂x ¯ ° ∂f ¯ ° ∂y

∂f ∂x

¶2

µ +

∂f ∂y

¶2 .

 s µ ¶2 µ ¶2     ∂f ∂f    g dS =  + dxdy. g(x, y, f (x, y)) 1 +     ∂x ∂y S D

By the first remark, setting g(x, y, z) = 1 yeilds our previous formula for the area of that part of the surface S given z = f (x, y) lying over the region D in the x-y plane, namely:   s µ ¶2 µ ¶2     ∂f ∂f    + dxdy. 1 + A(S) =      ∂x ∂y D

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7. Vector Fields So far we have concentrated on developing the calculus of scalar valued functions of two or more variables. Henceforth our focus will be on vector valued functions. A vector field in R2 is a vector valued function F : R2 → R2 assigning to each point (x, y) the vector F(x, y) = (P (x, y), Q(x, y)) = P (x, y)i + Q(x, y)j, where P and Q are the scalar valued component functions of F. For example, F(x, y) = −yi + xj, see figure 37. y

0

x

Figure 37. The vector field F(x, y) = −yi + xj Similarly, a vector field in R3 is a vector valued function F : R3 → R3 assigning to each point (x, y, z) the vector F(x, y, z) = (P (x, y, z), Q(x, y, z), R(x, y, z)) = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k. See figure 38. Such vector fields may for example represent: (1) A gravitational field, where F(x, y, z) is the gravitational force acting on a unit mass located at the point (x, y, z) due to a distribution of masses in space. (2) An electrical field, where F(x, y, z) is the force acting on a unit positive charge located at the point (x, y, z) due to a distribution of electric charges in space. (3) A velocity field, where for instance F(x, y, z) represents the velocity (speed and direction) at the point (x, y, z) of a steadily flowing fluid. Alternatively, for a planar fluid flow, F(x, y, z) = (P (x, y, z), Q(x, y, z), 0) could represent the instantaneous velocity of the fluid at the point (x, y) at time z.

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MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA z

F(x, y, z)

(x, y, z)

y

x

Figure 38. The vector field F(x, y) = −yi + xj

Given a scalar valued function f (x, y, z) its gradient, grad f = ∇f =

∂f ∂f ∂f i+ j+ k ∂x ∂y ∂z

defines a vector field in R3 . If a vector field F(x, y, z) arrises in this way; that is, F(x, y, z) = ∇f =

∂f ∂f ∂f i+ j+ k, ∂x ∂y ∂z

for some scalar function f , we refer to F as a gradient (or conservative) vector field. The underlying function f (x, y, z) is often referred to as a potential function for the vector field F(x, y, z). For example, GM GM f (x, y, z) = p = krk x2 + y 2 + z 2 is a (gravitational) potential function for the gravitational field; F(x, y, z) =

−GM x (x2 + y 2 + z 2 ) +

=

3 2

i+

−GM z 3

(x2 + y 2 + z 2 ) 2

−GM y 3

(x2 + y 2 + z 2 ) 2

j

k

−GM r , krk2 krk

of a point mass M located at the origin. As we know from before a gradient field F(r) at a point r = (x, y, z) is perpendicular to the level curve/surface of the potential function through that point.

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57

Given a vector field F(x, y, z) = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k, we can use the vector differential operator ∇ := i

∂ ∂ ∂ +j +k ∂x ∂y ∂z

to define two associated quantities. The divergence of F is defined to be: div F := ∇ · F =

∂P ∂Q ∂R + + . ∂x ∂y ∂z

The divergence, div F, is a scalar valued function and, as will become clearer later, when F is the velocity field for a fluid flow div F(x, y, z) represents the net rate of flow of fluid out of the point (x, y, z); that is, the tendency of the fluid to ‘diverge’ from the point. If at all points div F = 0 then we say the vector field F is incompressible. The curl of F is defined to be: curl F := ∇ × F ¯ ¯i ¯∂ = ¯¯ ∂x ¯P µ =

∂ ∂y

Q

¯ k ¯¯ ∂ ¯ ∂z ¯ R¯

∂R ∂Q − ∂y ∂z µ

+

j

¶

∂Q ∂P − ∂x ∂y

µ i+

∂P ∂R − ∂z ∂x

¶ k

For a vector field F(x, y) = P (x, y)i + Q(x, y)j in R2 , div F := ∇ · F =

∂P ∂Q + . ∂x ∂y

If we identify F(x, y) with the vector field F(x, y, z) := (P (x, y), Q(x, y), 0) defined on the x-y plane in R3 , then it makes sense to define curl F := ∇ × F ¯ ¯i ¯ ∂ ∂j = ¯¯ ∂x ∂y ¯P Q µ ∂Q = − ∂x

¯ k ¯¯ ∂ ¯ ∂z ¯ 0¯ ¶ ∂P k. ∂y

¶ j

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Again for reasons that will become apparent later, when F is the velocity field for a fluid flow, curl F measures the tendency of the fluid to rotate about each point. Points near the point (x, y, z) tend to rotate about an axis through (x, y, z) parallel to curl F(x, y, z) and with an angular velocity related to kcurl F(x, y, z)k. If everywhere curl F = 0 we say the vector field F is irrotational. Theorem 7.1. If f (x, y, z) has continuous second partial derivatives, then curl grad f = ∇ × ∇f = 0. Proof.

¯ ¯i ¯∂ ∇ × ∇f = ¯¯ ∂x ¯ ∂f ∂x

j ∂ ∂y ∂f ∂y

¯ k¯ ∂ ¯¯ ∂z ¯ ∂f ¯ ∂z

µ

¶ ∂ 2f ∂ 2f = − i + ··· ∂y∂z ∂z∂y = 0i + 0j + 0k = 0. Thus, provided the potential function has continuous second order partial derivatives, a conservative vector field is irrotational. This gives us a necessary condition for a vector field F to be conservative; if curl F is not identically zero, then F cannot be expressed as the gradient of any potential function. Thus, for example the vector field F(x, y) = −yi + xj, for which curl F = 2, cannot be written as the gradient of any potential function; i.e., is not a conservative vector field. We will later see that the converse of this is also true: if F is a vector field whose domain is simply connected (that is, ‘has no holes’), whose component functions have continuous partial derivatives and curl F = 0, then F is a conservative vector field. Similarly we have:Theorem 7.2. If F(x, y, z) = P i + Qj + Rk is a vector field on R3 and P , Q and R have continuous second partial derivatives, then div curl F = ∇ · ∇ × F = 0. Proof.

µ ¶ µ ¶ ∂ ∂R ∂Q ∂R ∂ ∂P ∇·∇×F= − − + ∂x ∂y ∂z ∂y ∂z ∂x µ ¶ ∂ ∂Q ∂P + − ∂z ∂x ∂y ∂2Q ∂ 2P ∂ 2R − + = ∂x∂y ∂x∂z ∂y∂z ∂ 2Q ∂2P ∂ 2R + − − ∂y∂x ∂z∂x ∂z∂y = 0,

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59

as the order of taking second partial derivatives commutes. Thus, if a vector field F is not incompressible; i.e., div F 6= 0, then F cannot be written as the curl of another vector field; i.e., F 6= ∇ × G for any field G. For example: For F = r = xi + yj + zk, we have ∇ · F = 3, so F is not purely rotational; that is F 6= curl G for any G. A differential operator that arises frequently in engineering and the physical sciences is Laplace’s operator: ∇2 := ∇ · ∇; that is, for a scalar valued function f (x, y, z), ∇2 f = ∇ · (∇f ) = div grad f µ ¶ µ ¶ ∂f ∂ ∂ ∂ ∂f ∂f = i +j +k · i+ j+ k ∂x ∂y ∂z ∂x ∂y ∂z ∂ 2f ∂ 2f ∂2f = + + . ∂x2 ∂y 2 ∂z 2 So, ∇

2

∂2 ∂2 ∂2 = + + . ∂x2 ∂y 2 ∂z 2

For a vector field F = P i + Qj + Rk we take ∇2 F := ∇2 P j + ∇2 Qj + ∇2 Rk. NOTE: We have effectively shown that the potential function g = GM/r for the gravitational field of a massive particle at the origin satisfies Laplaces equation: ∇2 g = 0. This is true for the potential function for the gravitational field of any distribution of masses, and is an important partial differential equation which arises in many contexts. There are a large number of identities relating the operators grad, div, curl and ∇2 to one another; for example, ∇ × ∇ × F = ∇(∇ · F) − ∇2 F. That is, curl curl F = grad div F − ∇2 F. For a more complete list, see Stewart (Fourth Edition), section 17.5, Exercises 23 to 29, on page 1116. Some applications of these are to be found in the subsequent exercises; 30, 31 and 32.

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7.1. Line integrals of vector fields. Suppose F = P i + Qj + Rk is a continuous vector field and r(t) = x(t)i + y(t)j + z(t)k, for a ≤ t ≤ b, is a parameterized curve C in R3 , or R2 (in which case we would have the x-y plane as the domain of F with P and Q depending only on x, and y, R = 0 and z(t) = 0). If T(t) is the unit tangent vector to the curve C at the point r(t) pointing in the direction of increasing t, then we know that r0 (t) T(t) = 0 . kr (t)k and at points of C we can form the scalar valued function F(r(t)) · T(t) representing the tangential component of F to the curve at r(t), see figure 39. z F(r(t)) T(t)

0 y

x

Figure 39. The vectors F(r(t)) and T(t). We then have, Z Z b r0 (t) F · T ds = F(r(t)) · 0 kr0 (t)kdt kr (t)k C a Z b = F(r(t)) · r0 (t)dt a Z b = (P i + Qj + Rk) · (x0 (t)i + y 0 (t)j + z 0 (t)k) dt a Z b = (P ((x(t), y(t), z(t))x0 (t) a

+ Q((x(t), y(t), z(t))y 0 (t) + R((x(t), y(t), z(t))z 0 (t)) dt. These right hand side integrals are often abreviated as Z Z F · dr, or P dx + Qdy + Rdz, C

C 0

using the convention dx = x (t)dt etc.

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61

R When F is a gravitational (electric, . . . ) force field, C F · dr represents the work done by the field on a unit mass (charge, . . . ) as it is moved from r(a) to r(b) along C (equals the energy gained by the particle). Fundamental theorem for line integrals By analogy with the Fundamental Theorem of Calculus (and a direct consequence of it) we have:Theorem 7.3. Let C be a smooth curve given parametrically by r = r(t), for a ≤ t ≤ b, and let f be a function of two or three variables for which grad f exists and is continuous on C, then Z ∇f · dr = f (r(b)) − f (r(a)). C

Proof. . . . , see Stewart (fourth edition), section 17.3, Theorem 2. Corollary 7.4. If C is a smooth curve given parametrically by r = r(t), for a ≤ t ≤ b, and F is a conservative vector field that is continuous at the points of C, then Z F · dr = f (r(b)) − f (r(a)), C

where f is a potential function for F. Corollary 7.5. If F is a continuous conservative vector field and C1 and C2 are two smooth curves lying entirely in the domain of F and with the same initial and terminal points, then Z Z F · dr = F · dr C1

C2

That is, line integrals of a continuous conservative vector field are ‘path independent’. Theorem 7.6. The line integrals of a continuous vector field F are path independent if and only if Z F · dr = 0 C

for every closed smooth curve C lying in the domain of F. Proof. . . . , see Stewart (fourth edition), section 17.3, Theorem 3. Thus, for a conservative continuous vector field F, Z F · dr = 0 C

for every closed smooth curve C lying in the domain of F. Under additional assumptions on the domain of F, the converse of this is also true:

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MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

Theorem 7.7. If F is a continuous vector field on an open and path-wise connected domain, D, for which Z F · dr = 0. C

whenever C is a closed smooth curve lying in the domain of F, then F is a conservative vector field.

NOTE: The proof in Stewart (fourth edition) section 17.3, Theorem 4, pp 1095-96 is incorrect. The correct proof is given here. Proof. (For simplicity, we will restrict ourselves to the case of vector fields on R2 ) Choosing any (a, b) ∈ D, for any point (x, y) ∈ D ∃ a path C, from (a, b) to (x, y) and R F · dr is independent of which path is chosen. That is, it depends only on (x, y), so C we define the function

Z

(x,y)

F · dr

f (x, y) = (a,b)

If ∆x is small enough then (x + ∆x, y) ∈ D (as D is open). Let C1 be a horizontal line segment from (x, y) to (x + ∆x, y). See figure 40.

(x,y)

C1 (x + ∆x, y)

C

D (a,b)

Figure 40. The curve C and the horizontal line segment C1 .

MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

Z

Z

f (x + ∆x, y) − f (x, y) =

F · dr − Z

63

C∪C1

=

F · dr C

F · dr C1 Z (x+∆x)

=

[P (x, y)dx + Q(x, y)dy] Z

x (x+∆x)

=

P (x, y)dx

(since dy = 0 on C1 )

x

= P (x + ξ, y)∆x

for some ξ ∈ [0, ∆x],

by the intermediate value theorem. So, taking the limit as ∆ → 0 we have

A Similar argument yields

∂f = P (x, y). ∂x ∂f = Q(x, y). ∂y

So ∂f ∂f i+ j ∂x ∂y = ∇f

F=

and so F is conservative. ¤ . If F = P i+Qj is a conservative vector field for which P and Q have continuous partial derivatives, then for f any potential function for F we have ∂f ∂f P = andQ = ∂x ∂y and so ∂P ∂ 2f ∂ 2f ∂Q = = = ∂y ∂y∂x ∂x∂y ∂x That is, for a conservative vector field F = P i + Qj, for which P and Q have continuous partial derivatives, we have ∂Q ∂P = . ∂y ∂x

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MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

For a vector field F, whose domain is open and simply connected (path-wise connected with no holes), the converse is also true. This is a consequence of the following important theorem. Green’s theorem: Theorem 7.8. Let C be an anticlockwise oriented simple closed curve in R2 , let D be the region bounded by C, and let F = P i + Qj be a vector field for which P and Q have continuous partial derivatives throughout an open region containing D, then Z ZZ F · dr = (∇ × F) · k dA. C

Or, equivalently

D

ZZ µ

Z P dx + Qdy = C

D

∂Q ∂P − ∂x ∂y

Proof. . . . , see Stewart (fourth edition), section 17.4.

¶ dxdy.

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65

8. Surface Integrals of Vector Fields Let S be a bounded surface in R3 given parametrically by r(u, v) = (x(u, v), y(u, v), z(u, v)), for (u, v) ranging over some bounded domain D in the u-v plane, and for which the component functions x(u, v), y(u, v) and z(u, v) have continuous partial derivatives. Let F(x, y, z) be a continuous vector field whose domain contains the surface S. The surface integral over S of the normal component of F to the surface is ZZ

ZZ F · dS := S

F · n dS, S

where n :=

ru × rv , kru × rv k

is the unit normal to the surface at the point specified parametrically by (u, v). Here it is assumed that ru × rv is never zero on S. Thus S is orientable; the normal always points to one side of S. Note: We could also choose as our normal direction rv × ru in which case the value of the surface integral would be the negative of the one given here. When F is the velocity field for a steady fluid flow this integral represents the net flow (or flux) across the surface S in the direction of the normal. From our work on surface integrals we have: ZZ

ZZ F · dS := S

F(r(u, v)) · n dS ZZ

D

=

F(r(u, v)) · Z ZD

=

ru × rv kru × rv k dudv kru × rv k

F(r(u, v)) · (ru × rv ) dudv. D

When S is given by z = f (x, y) for (x, y) ∈ R and F = P i + Qj + Rk, ZZ F · dS S

reduces to, ZZ R−P R

∂f ∂f −Q dxdy. ∂x ∂y

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MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

8.1. Stokes’ Theorem. Stokes’ theorem is a more general form of Green’s theorem. Theorem 8.1.

Z

ZZ F · dr =

∇ × F · dS,

C

S

Here C is the boundary of S oriented according to the right hand rule: if the thumb of the right hand points in the direction of the normal n to the surface, then the fingers of the right hand point in the direction of positive orientation for C (See figure 41). z n

S C 0 y x

Figure 41. The boundary of a surface oriented according to the right hand rule. Here it is assumed that the component functions of F have continuous partial derivatives on an open domain containg S and that S is an oriented piecewise-smooth surface whose boudary, C, is a simple closed piecewise-smooth curve. P roof. . . . , see Stewart (fourth edition), section 17.8. Corollary 8.2. If S is a closed oriented surface, then ZZ ∇ × F · dS = 0. S

Proof. Punch a small R hole in the surface and note that if C denotes the closed boundary of the hole. Then C F · dr → 0 as the hole gets smaller. ¤ The Divergence (Gauss’) Theorem Under appropriate assumptions, if D is the region bounded by an anticlockwise oriented simple closed curve C in R2 , then for a vector field F = P i + Qj defined on an open region containing D, Green’s theorem provides an expression for the line integral of the tangential component of F about C, namely; Z ZZ F · dr = (∇ × F) · k dA, C

D

MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

or equivalently,

67

ZZ µ

¶ ∂Q ∂P P dx + Qdy = − dxdy. ∂x ∂y C D This may be applied to also obtain an expression (sometimes known as Green’s second theorem) for Z Z Z

F · ds := C

F · n ds, C

the line integral of the normal component of F about C. NOTE: here, ds is shorhand for the ‘infinitesimal vector’ n ds. When F is the velocity field for a steady planar fluid flow, the integral Z F · ds C

represents the net flow (or flux) across the curve C. If C is given parametrically by, r(t) = x(t)i + y(t)j,

for a ≤ t ≤ b,

with r(b) = r(a), then r0 (t) = x0 (t)i + y 0 (t)j is tangential to C at the point specified by t, and so the unit normal to C at this point is y 0 (t)i − x0 (t)j n(t) = . kr0 (t)k Thus, Z Z b F · n ds = (P (x(t), y(t))i + Q(x(t), y(t))j)· C

a

Z

y 0 (t)i − x0 (t)j 0 kr (t)k dt kr0 (t)k b

P (x(t), y(t))y 0 (t) − Q(x(t), y(t))x0 (t) dt Za Z = P dy − Qdx = (−Q)dx + P dy C C ZZ ∂P ∂Q = + dxdy, by Green’s theorem ∂y D ∂x applied to the field − Qi + P j, ZZ = ∇ · F dA. =

D

That is;

Z

ZZ F · ds = C

div F dA. D

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MULTIVARIABLE CALCULUS, UNIVERSITY OF NEWCASTLE, AUSTRALIA

When F is the velocity field for a planar steady fluid flow in R2 , this equates the net flow (or flux) across the curve C with the integral of the divergence of F over the planar region D bounded by C and expresses our interpretation of div F at a point as the rate per unit area at which fluid is ‘created’ at (flows out, or diverges from) that point. We might ask; is a similar result true for a simple closed surface S bounding a solid region R in R3 ? The answer is yes, and this is the divergence (or Gauss’s) theorem: Theorem 8.3. Let R be a solid region in R3 bounded by a smooth simple closed outward oriented surface S and let F be vector field whose component functions have continuous partial derivatives on an open region containing R, then ZZ ZZZ F · dS = div F dV. S

R

Proof. . . . , see Stewart (fourth edition), section 17.9. NOTE: When F represents the velocity field of a steady fluid flow in R3 , the left hand integral represents the net flow (flux) of fluid out of the region R across its boundary S. Gauss’s theorem may be seen as equating this with the average value of the divergence of F in R multiplied by the volume of R; that is, the net amount of fluid being ‘created’ (diverging from points) within R. From the divergence theorem and the identity div curl F = 0 we can reestablish corollary 8.2: For S a simple closed oriented surface and F a vector field whose component functions have continuous partial derivatives on an open region containing the solid region R bounded by S ZZ ZZZ ∇ × F · dS = S

div curl F dV = 0. R